0.5 N solution of KCl is isotonic with ..... | vedclass

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(C) Two solutions are isotonic if their osmotic pressures are equal,i.e.,$\pi_1 = \pi_2$. For a non-electrolyte like sugar,the osmotic pressure is $\p

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(C) Two solutions are isotonic if their osmotic pressures are equal,i.e.,$\pi_1 = \pi_2$. For a non-electrolyte like sugar,the osmotic pressure is $\pi = C 	imes R 	imes T$. For an electrolyte like $KCl$,the osmotic pressure is $\pi = i 	imes C 	imes R 	imes T$,where $i$ is the van't Hoff factor. Since $KCl$ dissociates as $KCl ightarrow K^+ + Cl^-$,the number of ions produced is $2$,so $i = 2$. Given the normality $(N)$ of $KCl$ is $0.5 \ N$. Since the n-factor for $KCl$ is $1$,the molarity $(M)$ is equal to the normality $(N)$,so $M = 0.5 \ M$. Equating the osmotic pressures: $i_{KCl} 	imes M_{KCl} = M_{sugar}$. $2 	imes 0.5 = M_{sugar}$. $M_{sugar} = 1 \ M$. Since the n-factor for sugar is $1$,the normality of the sugar solution is $1 \ N$.

$0.5 \ N$ solution of $KCl$ is isotonic with $.........$ solution of sugar,assuming $100\%$ dissociation for $KCl$.

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